The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain


 Millicent Martin
 1 years ago
 Views:
Transcription
1 Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain (d) phase margin or (e) gain margin  in frequency domain. If the given plant does not satisfy 2 or 3 conditions specified from the above, a compensator circuit may be necessary to modify the overall plant characteristics (i.e of G c(s) G f(s)) to satisfy the given specifications. The controllers used are typically (other than P, I, D combination controllers) Lead compensator (to speed up transient response, margin of stability and improve error constant in a limited way) Lag compensator (to improve error constant or steadystate behavior while retaining transient response) Lead Lag compensator (A combination of the above two i.e. to improve steady state as well as transient). Lead Compensator: is so called since it adds phase angle so that the phase curve of the combined system is lifted above the line so much as to provide the necessary phase margin. A compensator 1 1 gives an angle Ф c. If 1, then Ф c is positive. (or lead). In pole  zero form, it is given as G c = = = with = Z c/ P c also gives a lead angle. 1
2 Lead compensator can be realized with electrical or mechanical components G c(s) Letting = R 1 C; and = R 2 /(R 1 +R 2) < 1 G c(s) = / / = =  (1) with τ = B 1 / K 1 α = K 1 /( K 1 + K 2) Y(s)/X(s) = + 1) / 1) The phase difference for G c(s) given by eqn. (1). is Ø c or tan Ф = 1 /1 dф /d = 0 gives the frequency for maximum phase difference as ω  which is the geometric mean of corner frequencies. The max. phase lead is given as tan Ф or Ф Ф The maximum gain is 20 log 1/ at high frequency, and the gain at ω G c(ω = = or log G c(ω =10 log, db is the gain contribution by compensator at the mean frequency. or sin Ф 2
3 Design of Compensator for a system Design can be carried on in frequency domain or in time domain using Rootlocus Both methods are explained for lead & lag compensator cases. Let us consider Frequency domain approach first in each case. Lead Compensator : Design procedure steps For the given plant determine K v or K a based on specification Draw Bode Plot and check whether adequate Ф is provided If is specified, find the required phase margin from approx relation 0.01 Ф margin Or Ф margin = becomes the specified phase margin Determine the phase lead to be introduced by controller from Ф e = Ф s Ф uc + Where Ф s is the specified phase margin, Ф uc  phase margin of uncompensated system, and to account for shift of crossover frequency. Let Ф m = Ф l, then determine from Ф Ф (If Ø m is more than 60 0, it is desirable to provide two identical networks each providing ½ the phase) Draw a line 10 log1/ below 0, db on the uncompensated Bode plot and mark the corresponding frequency 1 = 1/ = m ; 2 = 1/ = m / ; G c(s) = (= / / ) Compensated system is G c(s) G f(s) G c G f Example (from Nagrath & Gopal) Lead Compensator in frequency domain To design a compensator for a system G f(s) = so that Acceleration Error Const. Ka = 10, and phase margin = 350. K a = s 2 G(s) s = K = 10. Hence choose K= 10 Draw Bode Plot for G f(s) =.. Phase margin is 33 0, system quite unstable. Phase lead to be introduced by network = 35 0 (33 0 ) +. 3
4 Chose 12 to 14 0 as gain slope in high. Thus Ф e = = Being high, two networks in tandem are required With Ф m = (1sin 41 0 )/(1+sin 41 0 ) = log 1/ for lead network occurs at the frequency where Ф is maximum. Hence check the frequency c2 where the gain of UC system is 10 log 1/ (actually 2 x 10log 1/ in this case as 2 networks are used). c2 = 6.3 rad/sec (at db) = m c2 = 6.3 rad/sec should be designed as the mean frequency ( m ) Thus 1 = m = 2.8 r/sec ; 2 = m = 14.0 r/sec ; Thus G e(s) =.. =.. and the compensated plant is (
5 Lag Compensator can be realized as shown Letting = R 2 C ; = (R 1 + R 2) /R 2 > 1 The transfer function becomes G c(s) = ( (also 1 ) Ø c which is ve as >1 Hence the contribution to phase angle of G f(c) is negative where ρ = (K 1+K 2)/K 1 >1 Note: Relations for Ф max (at the geometric mean of = & = ) etc. are similar to lead compensator. However, lag (or decreasing the phase difference further is undesirable). The characteristic taken advantage is high frequency gain attenuation (reduction by 20log decibels) with very little decrease in phase. 5
6 Lag Compensator Design procedure This can be used only when there is a range of Ф more than in G f(s), since phase lag increases due to lag compensator. The Compensator relies on reducing the gain of the compensated plant in the critical region to make it adequately stable. Find the openloop gain necessary to satisfy the required condition Find the frequency ω c2 when uncompensated system makes Ø 2 =Ø s + where is 5 0 to 15 0 to account for phase lag due to network Measure the gain of UC system at ω c2, and equate it to 20 log and find. Choose upper corner frequency ω 2 (= 1/) of network one octave or decade below ω c2, (ω 2 = ω c2 /2 ie. Octave or ω c2 /10 for decade) Knowing &, ω 2 (= 1/); w 1 (= 1/); G c = ( Following example will clarify: Example: Design a lag compensator so that the plant G f(s) = the specifications: ξ = 0.4, t s = 1.0 sec; K v= 5; meets Solution: K v = s G(s) s o = k/4 => K=20 ; Using approximation Ø s = ξ/0.01 = 0.4/0.01 = 40 0 (Bode plot for the system) G f(s) = is as shown. When amplitude ratio, in log terms G f(jω) is zero, the Ф curve is 4 0 below , Ф 2 < Hence the system is unstable. To the required Ф s = 40 0, add = 10 0 to compensate for negative Ф s. Ф = =
7 Frequency where Ф = is 0.6 rad/sec (ω c2 ) choose ω 2 at 2 octaves less than ω c2 ω 2 = 0.6/2 2 = 0.15 rad/sec Also the gain at A has to be brought down to 0 when Ф = Hence 20 log = 19, or = = 1/ = 0.15/8.91 = Hence the compensator to be implemented is G c(s) = ( =... The compensated plant is G c(s) G f(s)type equation here. = Compensator Design Root Locus Approach (Timedomain) Lead Compensator design procedure:: Translate the desired specifications into a dominant pair of complex poles in the Root Locus. (complex plane). If the given system G f(s) is such that G f(s) at S d then a compensator needs to be designed such that G f(s) G c(s) at Sd or G c(s) d G f(s) d = Φ (A) ie. the polezero pair of compensator have to contribute Φ. From the above figure, we have from properties of s, z c / n = Φ Φ So that zc /pc = and pc/n = Φ Φ Φ and d/d = z c /p = sin 7
8 (to reduce the effect of gain due to compensator, it is desirable to have large ). which leads to = ½ ( Φ (B) Thus after locating Sd in the complex plane, knowing = Cos 1 ξ, and Ø from (A), locate z c from (B), as well as p c Knowing z c and p c, the compensator is (s+z c)/(s+p c) Example Lead Compensator design  Time domain Design a compensator is give ξ = 0.5 and n = 2 for the system G f(s) = Solution: Given the damping ratio ξ = 0.5(=Cos 1 ) and n of the dominant complex pole pair, the point S d to pass through the root locus of the compensated plant is Sd=  ξ n j n1 ξ 2 ) = 1 j 1.73 = Cos = 60 0 Φ = 180 G f(s d) = ( ) = 60 0 ν= ½ ( Φ) = ( 60 60/2= 30 0 However, locating oz so that ν =30 0 makes Z c coincide with the pole of (1,0) of the given system. Hence choose Z c = 1.2 close to 1 Now draw OP c so that angle Z c O P c = Φ = 60 0 P c cuts the real axis at so that G c =. =. Also the gain at Sd(= 1+ j 1.73) for the compensated system is K = s. s+1. s+4. s / s+1.2 ==30, in which s, s+1 etc are the lengths of vectors which can be measured and substituted. Hence compensated system = G f G c = 30(s +1.2) / s(s+1) (s+4) (s+4.95) 8
9 Lag Compensator Time Domain Method: Lag compensator is designed when the steady state response needs improvement, while transient performance is satisfactory. As the transient performance is satisfactory, the technique lies in ensuring that the root locus passes through the same Sd as for the uncompensated system while the error constant is improved. Therefore pole P is chosen close to origin and zero to its left so that β =Z c /P c >1 where Φ = PSZ is 2 to 5 0 and OSZ < 10 0, We can establish value of β using the following reasoning: Gain of UC system at Sd = K uc.... (Sd) =. Similarly K c (Sd) =..... X Since a b, we have K uc (S d) = K c (S d) Now, Error Constant K e c = K c (Sd) x = K uc (Sd) x Ke uc. Hence β = = Ke c /Ke uc Thus β can be chosen as the ratio of the expected Error Constant to the Error Constant of the given system 9
10 Example: Design a lag compensator for the system G f (s) = to meet the following specs: Damping ratio = ζ = 0.5; settling time t s =10 secs & K v 5 sec 1 Solution: From t s = 10 = 4/ζ n n = 4/ 10 x 0.5 = 0.8 Thus Sd =  ζ n + j 1 ζ 2 n = j 0.7 Sd lies on the uncompensated system as well. At Sd, gain of the UC system is K uc = OS x SA x SB = 0.8 x 0.9 x 3.7 = 2.66 Velocity Error Constant of UC system K uc v = s G f (s) = 2.66/(1 x 4) = so Thus β = K v desired / K v existing = 5/0.666 = 7.5 Say β = 10 to counter the effect of negative phase difference due to lag compensator. Now to locate the pole and zero, Draw a line from S to Z c so that OSZ c = 6 0. (small value) Determine OZ e = 0.1 (measure) Since β = Z c/p c, P c = Z/10 = 0.1/10 = 0.01 Also as the gain compensated system may be treated as the same as of UC system in which case.. G f G c =, which is the Compensated plant transfer function.... Or get K = = 2.2 instead of 2.66, by measuring the line lengths from the figure. ** ** ** 10
ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationFrequency Response Analysis
Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More information6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson
Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closedloop behavior what we want it to be. To review:  G c (s) G(s) H(s) you are here! plant For
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationPD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closedloop
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More informationFeedback design for the Buck Converter
Feedback design for the Buck Converter Portland State University Department of Electrical and Computer Engineering Portland, Oregon, USA December 30, 2009 Abstract In this paper we explore two compensation
More informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationFrequency (rad/s)
. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer
More informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationECE137B Final Exam. There are 5 problems on this exam and you have 3 hours There are pages 119 in the exam: please make sure all are there.
ECE37B Final Exam There are 5 problems on this exam and you have 3 hours There are pages 9 in the exam: please make sure all are there. Do not open this exam until told to do so Show all work: Credit
More informationThe FrequencyResponse
6 The FrequencyResponse Design Method A Perspective on the FrequencyResponse Design Method The design of feedback control systems in industry is probably accomplished using frequencyresponse methods
More informationDesign via Root Locus
Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steadystate error (Sections
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8 am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
More informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
More informationEE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More information2.010 Fall 2000 Solution of Homework Assignment 8
2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)
More information9/9/2011 Classical Control 1
MM11 Root Locus Design Method Reading material: FC pp.270328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequencydomain analysis and control design (15 pt) Given is a
More informationPM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
More informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationHomework 11 Solution  AME 30315, Spring 2015
1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
More informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
More informationLecture 7:Time Response PoleZero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 FirstOrder Specs: Step : Pole Real inputs contain
More informationChapter 10 Feedback. PART C: Stability and Compensation
1 Chapter 10 Feedback PART C: Stability and Compensation Example: Noninverting Amplifier We are analyzing the two circuits (nmos diff pair or pmos diff pair) to realize this symbol: either of the circuits
More informationPower System Control
Power System Control Basic Control Engineering Prof. Wonhee Kim School of Energy Systems Engineering, ChungAng University 2 Contents Why feedback? System Modeling in Frequency Domain System Modeling in
More informationSteady State Frequency Response Using Bode Plots
School of Engineering Department of Electrical and Computer Engineering 332:224 Principles of Electrical Engineering II Laboratory Experiment 3 Steady State Frequency Response Using Bode Plots 1 Introduction
More informationLecture 13: Internal Model Principle and Repetitive Control
ME 233, UC Berkeley, Spring 2014 Xu Chen Lecture 13: Internal Model Principle and Repetitive Control Big picture review of integral control in PID design example: 0 Es) C s) Ds) + + P s) Y s) where P s)
More informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK V SEMESTER IC650 CONTROL SYSTEMS Regulation 203 Academic Year 207 8 Prepared
More informationLab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More informationGoals for today 2.004
Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationDistributed RealTime Control Systems
Distributed RealTime Control Systems Chapter 9 Discrete PID Control 1 Computer Control 2 Approximation of Continuous Time Controllers Design Strategy: Design a continuous time controller C c (s) and then
More informationDesired Bode plot shape
Desired Bode plot shape 0dB Want high gain Use PI or lag control Low freq ess, type High low freq gain for steady state tracking Low high freq gain for noise attenuation Sufficient PM near ω gc for stability
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationControl Systems 2. Lecture 4: Sensitivity function limits. Roy Smith
Control Systems 2 Lecture 4: Sensitivity function limits Roy Smith 2017314 4.1 Inputoutput controllability Control design questions: 1. How well can the plant be controlled? 2. What control structure
More informationStability and Frequency Compensation
類比電路設計 (3349)  2004 Stability and Frequency ompensation hingyuan Yang National hunghsing University Department of Electrical Engineering Overview Reading B Razavi hapter 0 Introduction In this lecture,
More informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
More informationEssence of the Root Locus Technique
Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general setup, namely for the case when the closedloop
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an openloop system (a system without feedbac) and a closedloop
More informationLecture 17 Date:
Lecture 17 Date: 27.10.2016 Feedback and Properties, Types of Feedback Amplifier Stability Gain and Phase Margin Modification Elements of Feedback System: (a) The feed forward amplifier [H(s)] ; (b) A
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =
ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.
More informationUniversity of Science and Technology, Sudan Department of Chemical Engineering.
ISO 91:28 Certified Volume 3, Issue 6, November 214 Design and Decoupling of Control System for a Continuous Stirred Tank Reactor (CSTR) Georgeous, N.B *1 and Gasmalseed, G.A, Abdalla, B.K (12) University
More informationControl System. Contents
Contents Chapter Topic Page Chapter Chapter Chapter3 Chapter4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More informationAnalysis and Design of Control Systems in the Time Domain
Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.
More informationStep Response for the Transfer Function of a Sensor
Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant
More informationPitch Rate CAS Design Project
Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationD G 2 H + + D 2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Final Exam May 21, 2007 180 minutes Johnson Ice Rink 1. This examination consists
More informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
More informationProportional, Integral & Derivative Control Design. Raktim Bhattacharya
AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University
More informationLinear System Theory
Linear System Theory  Laplace Transform Prof. Robert X. Gao Department of Mechanical Engineering University of Connecticut Storrs, CT 06269 Outline What we ve learned so far: Setting up Modeling Equations
More information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More informationLecture 50 Changing Closed Loop Dynamic Response with Feedback and Compensation
Lecture 50 Changing Closed Loop Dynamic Response with Feedback and Compensation 1 A. Closed Loop Transient Response Waveforms 1. Standard Quadratic T(s) Step Response a. Q > 1/2 Oscillatory decay to a
More informationSecond Order and Higher Order Systems
Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical secondorder control system to a step input. In terms of damping ratio and natural
More informationLaboratory III: Operational Amplifiers
Physics 33, Fall 2008 Lab III  Handout Laboratory III: Operational Amplifiers Introduction Operational amplifiers are one of the most useful building blocks of analog electronics. Ideally, an op amp would
More informationAutonomous Mobile Robot Design
Autonomous Mobile Robot Design Topic: Guidance and Control Introduction and PID Loops Dr. Kostas Alexis (CSE) Autonomous Robot Challenges How do I control where to go? Autonomous Mobile Robot Design Topic:
More informationFREQUENCYRESPONSE ANALYSIS
ECE450/550: Feedback Control Systems. 8 FREQUENCYRESPONSE ANALYSIS 8.: Motivation to study frequencyresponse methods Advantages and disadvantages to rootlocus design approach: ADVANTAGES: Good indicator
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationChapter 3 AUTOMATIC VOLTAGE CONTROL
Chapter 3 AUTOMATIC VOLTAGE CONTROL . INTRODUCTION TO EXCITATION SYSTEM The basic function of an excitation system is to provide direct current to the field winding of the synchronous generator. The excitation
More informationCDS 101: Lecture 8.2 Tools for PID & Loop Shaping
CDS : Lecture 8. Tools for PID & Loop Shapig Richard M. Murray 7 November 4 Goals: Show how to use loop shapig to achieve a performace specificatio Itroduce ew tools for loop shapig desig: ZieglerNichols,
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationReglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16
Reglerteknik Allmän Kurs Del Lösningar till Exempelsamling Läsår 5/6 Avdelningen för Reglerteknik, KTH, SE 44 Stockholm, SWEDEN AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More information